Ac Fundamental.docx

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There are numerous sources of DC and AC current and voltage. However:

DC Current vs. AC Current

 

Direct current (DC) flows in one direction the circuit. Alternating current (AC) flows first in one direction then in the opposite direction.

Sources of DC are commonly shown as a cell or

battery:

The same definitions apply to alternating voltage (AC voltage):  

Sources of AC are commonly shown as an AC

DC voltage has a fixed polarity. AC voltage switches polarity back and forth.

generator: Top The Sinusoidal AC Waveform The most common AC waveform is a sine (or sinusoidal) waveform.

The sine waveform is accurately represented by the sine function of plane trigonometry: y = rsin where:

The vertical axis represents the amplitude of the AC current or voltage, in amperes or volts. The horizontal axis represents the angular displacement of the waveform. The units can be degrees or radians.

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y = the instantaneous amplitude r = the maximum amplitude = the horizontal displacement

Instantaneous Current and Voltage

i = Ipsin where i = instantaneous current in amperes Ip = the maximum, or peak, current in amperes  = the angular displacement in degrees or radians

v = Vpsin where v = instantaneous voltage in volts Vp = the maximum, or peak, voltage in volts  = the angular displacement in degrees or radians

Topic 1-2.1 Peak and Peak-to-Peak Voltage Peak and Peak-to-Peak Voltage

Peak voltage is the voltage measured from the baseline of an ac waveform to its maximum, or peak, level. Unit: Volts peak (Vp) Symbol: Vp

Peak and peak-to-peak values are most often used when measuring the amplitude of ac waveforms directly from an oscilloscope display.

Peak-to-peak voltage is the voltage measured from the maximum positive level to the maximum negative level. Unit: Volts peak-to-peak (Vp-p) Symbol: Vp-p

Convert Vp to Vp-p:

For a typical sinusoidal waveform, the positive peak voltage is equal to the negative peak voltage. Peak voltages are expressed without a + or - sign.

For a typical sinusoidal waveform, the peak-topeak voltage is equal to 2 times the peak voltage. Peak-to-peak voltages are expressed without a + or - sign .

What is the peak-to-peak value of a sinusoidal waveform that has a peak value of 12 V?

Vp-p = 2 Vp Ans: 24 Vp-p Convert Vp-p to Vp: Vp =0.5Vp-

What is the peak value of a sine wave that has a peak-to-peak value of 440 V?

p

Ans: 220 Vp

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Topic 1-2.2 RMS Voltage Root-Mean-Square (RMS) Voltage

RMS voltage is the amount of dc voltage that is required for producing the same amount of power as the ac waveform. Unit: Volts (V) Symbol: Vrms The RMS voltage of a sinusoidal waveform is equal to 0.707 times its peak value.

AC levels are assumed to be expressed as RMS values unless clearly specified otherwise.

In a dc circuit, applying 2 V to a 1 resistance produces 4 W of power. In an ac circuit, applying 2 Vrms to a 1  resistance produces 4 W of power. RMS voltages are expressed without a + or - sign.

Vrms = 0.707Vp

Convert Vp to Vrms: Vrms = 0.707Vp Convert Vrms to Vp

:

Determine the RMS value of a waveform that measures 15 Vp. Ans: 10.6 V Determine the peak value of 120 V. (Hint: Assume 120 V is in RMS)

Vp =1.414Vrms

Ans: 170 Vp

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Topic 1-2.3 Average Voltage Average Voltage

Average voltage is the average value of all the values for one half-cycle of the waveform.

(Vave)

The average voltage is determined from just one half-cycle of the waveform because the average value of a full cycle is

Symbol: Vave

zero.

The average voltage of a sinusoidal waveform is equal to 0.637 times

Average voltages are expressed without a + or sign

Unit: Volts average

its peak value. Vave = 0.637Vp

Convert Vp to Vave: Determine the average value of a waveform that measured 16 Vp. Vave = 0.637Vp

Ans: 10.2 Vave What is the peak value of a waveform that has an average value of 22.4 V?

Convert Vave to Vp: Ans: 35.1 Vp Vp =1.57Vave

One period occupies exactly 360º of a sine waveform.

Period of a Waveform The period of a waveform is the time required

The usual units of measure are:

for completing one full cycle.

seconds (s) milliseconds (ms) microseconds (ms)

Math symbol: T Unit of measure: seconds (s)

The usual units of

Frequency of a Waveform

measure are:

hertz (Hz)

The frequency of a waveform is the number of cycles that is completed

kilohertz (kHz),

each second.

103 Hz

Math symbol: f Unit of measure: hertz (Hz)

megahertz (MHz), 106 Hz This example shows four cycles per second, or a waveform that has a frequency of 4 Hz.

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Period to Frequency f = 1/T

Frequency to Period T = 1/f

Frequency (f) is in hertz (Hz) Period (T) is in seconds (s)

gigahertz (GHz), 109 Hz

A certain sine waveform has a frequency of 100 Hz. What is the period of this waveform? Ans: 10 ms

What is the frequency of a waveform that has a period of 200 s? Ans: 5 kHz

Phase Angle

Two waveforms are said to be in phase when they have the same frequency and there is no phase difference between them.

The phase angle of a waveform is angular difference between two waveforms of the same frequency.  

Two waveforms are said to be out of phase when they have the same frequency and there is some amount of phase shift between them.

Math symbol:  (theta) Unit of measure: degrees or radians

Leading and Lagging Phase Angles A leading waveform is one that is ahead of a reference waveform of the same frequency. In this example, the blue waveform is taken as thereference because it begins at 0 degrees on the horizontal axis. The red waveform is said to be leading because it is already at about 90 degrees when the reference waveform begins at 0 degrees.

A lagging waveform is one that is behind a reference waveform of the same frequency.

In this example, the blue waveform is taken as thereference because it begins at 0 degrees on the horizontal axis. The red waveform is said to be lagging because it has not yet completed its cycle while the reference waveform is beginning a new one at 0 degrees.

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Keeping straight whether one waveform is leading or lagging another is commonly a confusing point for students of AC electricity (and no small number of practicing technicians as well). So it pays to keep in mind whatever pictures or gimmicks that are required for helping you specify which of two outof-phase waveforms is leading and which is lagging.

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Use the formula for instantaneous sine voltage and current to sketch an accurate sinusoidal waveform. Cite the basic equations and units of measure for DC power.

  

Note: User the BACK feature of your browswer to return to this page.

Sketch voltage, current, and power sine waveforms on the same axis. Explain why the power waveform is always positive as long as current and voltage are in phase. Cite the fact that average power is equal to the product of RMS current and RMS voltage.

AC Power Waveform

Notice that the power waveform is always positive. 

The current and voltagewaveforms are shown in phase. This is typical for a resistive load. The shaded green areas represent the corresponding levels ofpower.

The instantaneous value of power is equal to the instantaneous current times the instantaneous voltage. p = ie where:



A positive value of power indicates that the source is giving power to the load. A negative value of power would indicate that the circuit is returning power to the source (which will not happen in a resistor circuit).

The power waveform is always positive because the values of current and voltage always have the same sign-both negative or both positive. In algebra, this means that the product of the two values is always a positive value.

p = instantaneous value of power in watts i = instantaneous value of current in amperes v = instantaneous value of voltage in volts Average AC Power

It would seem more natural to say that the average power dissipation of a circuit is the product of the average values of current and voltage. But some simple math can show that it is not.

When the current and voltage waveforms are in phase, the average power is equal to the RMS voltage times the RMS current: Pave = IRMS x ERMS

Variations of the AC power equation include:

Conventional use allows us to write this equation more simply as: P = IE

I = P/E E=P/I

P = I2R P = E2 / R

It is then assumed that P is an average value and the other two terms are RMS values.

Before starting this module, you should be able to:

When you complete this module, you should be able to:



Describe the difference between period and frequency for a sinusoidal waveform.

   

Describe the features of a rectangular or squarewaveform. Calculate the total period and frequency of a rectangular waveform. Define duty cycle and average voltage as the terms apply to a rectangular waveform. Describe the features of a triangular or sawtoothwaveform.

Topic 1-6.1 Rectangular Waveform

Amplitude and Period A rectangular waveform is characterized by flat maximum and minimum levels, fastrising and fast-falling edges, and squared-off corners. Because of the squared corners, a rectangular waveform is also called a square waveform.

Rectangular waveforms are the heart of digital electronics. Virtually all digital waveforms have the characteristic on/off appearance. The frequency of a rectangular waveform is usually called itsrepetition rate rather than frequency.

The amplitude of a rectangular waveform is a measure of the distance between the minimum and maximum levels--the peak-to-peak value. Amplitude is most often expressed in units of volts, although units of current and power can be useful at times. The period of a rectangular waveform is the time required to complete one full cycle. The period is measured in units of seconds.

Period and Duty Cycle The period of a rectangular waveform can be further broken down into two phases:



The duty cycle of a waveform has no units -- it simply expresses the ratio of 'on' time to the total time. It is often expressed as a percentage where: duty cycle (%) = (TH / T ) x 100

Time High, TH -- The amount of time for the higher amplitude level. Time Low, TL--The amount of time for the lower amplitude level. In these terms, the period of the waveform can be give by:



Rectangular waveforms are sometimes used for regulating the amount of power applied to a load (such as a motor or lamp). The higher the duty cycle, the greater the amount of power applied to the load.



The frequency of a rectangular waveform is given by: f = 1 / T or f = 1/ (TH + TL)

However, there is no relationship between frequency and duty cycle of a rectangular waveform.

T = TH + TL T = total period of the waveform TH = time high TL = time low

Average Voltage

The duty cycle of a square waveform is the ratio of time high to the total period: duty cycle = TH / T

The average voltage of a square waveform is given by: Eave = Epeak x duty cycle

For a rectangual waveform, TL = 15 ms and TH= 10 ms. Calculate:

Step 1: To find the total period: T = TH + TL Step 2: To find the frequency: f = 1 / T

(a) The total period of the waveform. (b) The frequency of the waveform. (c) The duty cycle of the waveform.

Step 3: To find the duty cycle: duty cycle = TH / T

Ans: (a) 25 ms, (b) 40 Hz, (c) 0.4 or 40%

Top Topic 1-6.2 Sawtooth Waveform

Sawtooth Waveform A sawtooth waveform is characterized by one sloping edge and one that instantaneously returns to the baseline.

The slope of a sawtooth waveform is specified in terms of volts per second (V/s). If Vp is the amplitude and T is the period of sawtooth waveform, the slope is give by: slope = Vp / T Or since T = 1 / f: slope = Vpf

The amplitude of a sawtooth waveform is 12 V. If the frequency is 100 Hz, what is the slope?

slope = Vpf slope 12 V x 100 Hz slope = 1200 V/s

Lesson 2-1 Introduction to Inductance

Before starting this module, you should be able to:  

When you complete this module, you should be able to:

Describe a magnet field in terms of field strength, magnetic polarity, and flux density. Describe the characteristics of a magnetic field that are produced by electrons flowing through a conductor.

  

Faraday's Law for a Straight Wire

Express in words Faraday's Law for a straight wire. Cite the meaning of each term in the mathematical expression of Faraday's Law for a straight wire and for acoil of wire. Express Lenz's Law in a single sentence.

The faster the rate of change of flux, the larger the amount of induced voltage.

The amount of induced voltage is proportional to the rate of change of flux When there is no change in flux, there is no lines cutting the conductor. induced voltage.

where: Vind = the amount of induced voltage in volts (V) = the rate of change of flux cutting the conductor in webers/second (Wb/sec)

Faraday's Law for a Coil of Wire The amount of induced voltage is proportional to the rate of change of flux and the number of turns of wire.

Increasing the number of turns or the rate of change of flux increases the amount of induced voltage. Where: Vind = the amount of induced voltage in volts (V) N = the number of turns of wire = the rate of change of flux cutting the conductor in webers/second (Wb/sec)

Lenz's Law The voltage induced in a conductor will oppose the change in voltage that is causing the flux to change.

When an increasing voltage is applied to a conductor, the resulting increase in flux lines will induce a voltage that opposes the increase. When a decreasing voltage is applied to a conductor, the resulting decrease in flux lines will induce a voltage that opposes the decrease.

 

Cite the meaning of each term in the mathematical expression of Faraday's Law. Cite the meaning of each term in the mathematical expression of Lenz's Law.

Self-Inductance Self-inductance is the property of a circuit whereby a change in current causes a change in voltage.

   

Explain self-inductance. Define inductance in terms of induced voltage. Cite the units of measure for inductance. Describe the voltage waveform across an inductor when a sine waveform of current is being applied.

The amount of induced voltage (VL) is proportional to the rate of change of current flow (di/dt).   

where: VL = the induced voltage in volts, V L = the value of self-inductance in henries, H di / dt = the rate of change in current in amperes per second, A/T

When there is no change in current, di /dt = 0 and VL = 0. When the current is changing very rapidly, di/dt is a large value and so is the amount of VL. When the current is changing very slowly, di/dt is a small value and so is the amount of VL.

Self-inductance is also more simply known as inductance. The self-induction effect is multiplied by the value of inductance, L.  

Increasing the value of L increases the amount of voltage that is induced in response to a change in current. Decreasing the value of L decreases the amount of voltage that is induced in response to a change in current.

Inductance and Inductors Inductance is measured in units of Henries (H).

One henry is the amount of inductance that is required for generating one volt of induced voltage when the current is changing at the rate of one ampere per second.

The math symbol for inductance is L. VL = 1 V when L = 1 H and di /dt = 1 A / sec

The graphical symbol for an inductor resembles a coil of wire: Commonly used engineering units for inductance are:   

1 H = 1 henry 1 x 10-3 H = 1 mH or millihenry 1 x 10-6 H = 1 H or microhenry

Top Inductive Sine Waveforms 

R E A S O N I N G

 

The amount of induced voltage is proportional to the rate of change of current through an inductor. The value of a sine waveform is constantly changing. Therefore, when a sinusoidal current flows through an inductor (IL), the induced voltage (VL) changes constantly.

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efore starting this module, you should be able to:    

Explain self-inductance. Define inductance in terms of induced voltage. Cite the units of measure for inductance. Describe the voltage waveform across an inductor when a sine waveform of current is being applied.

When you complete this module, you should be able to:  

Define the term mutual inductance. Explain the meaning of coefficient of coupling.

MUTUAL INDUCTANCE Mutual inductance takes place between two coils. The amount of mutual inductance between these two coils depends on their respective values of inductance (L1 and L2) and the "strength" of the magnetic coupling between them (k).

Mutual inductance is the electrical property of circuits that enables a current flowing in one conductor (or coil) to induce a current in a nearby conductor (or coil).

where: M = mutual inductance in henries k = coefficient of coupling between two inductances L1 and L2 = values of the two inductances Unit of measure: henries (H) Math Symbol: LM or M

Coefficient of Coupling Two conductors, or coils, are said to be coupled when the are arranged so that a changing magnetic field created by one of the coils can induce a current in the other coil. The coefficient of coupling (k) between two coils indicates the degree of coupling between them. When k = 0, there is no coupling at all between coils This condition can occur, for example, when the coils are too far apart to interact or are separated by a The values of k are between 0 and 1 magnetic shield. -- the larger the value, the better the coupling. When k = 1, there is perfect coupling between the conductors or coils. There is no such thing as perfect coupling; however, a good transformer with an iron core will have a value of k that is close to 1.

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Lesson 3-2 Series Inductor Circuits

efore starting this module, you should be able to:  

When you complete this module, you should be able to:

Describe how inductors oppose any change in current flowing through them. Cite the units of measure for inductance.

    

Cite the equation for calculating the total inductance of a series circuit. Explain how total inductance increases with each inductor that is added in series. Calculate the total inductance of a series circuit, given the values of individual inductors. Describe how inductor voltage drops are distributed among inductors that are connected in series. Describe how changes in source voltage affect the current flowing through the circuit.

Topic 3-2.1 Equation for Total Series Inductance Total Inductance of a Series Inductor

A series inductor circuit consists of two or more inductors connected so that the total amount of circuit current flows through each inductor. The total inductance of a series inductor circuit is proportional to the values of the individual inductances.

Circuit The total inductance of a series inductor circuit is equal to the sum of the individual inductances: LT = L1 + L2 + L3 + ... + Ln where: LT = the total inductance in henries (H) L1, L2, L3, Ln = the value of the individual inductances in henries (H)

Determine the total inductance of this circuit when: L1 = 1 H L2 = 1.5 H Ans: 2.5 H

Determine the total inductance of this circuit when: L1 = 1 mH L2 = 150 H Ans: 1.15 mH

Determine the total inductance of this circuit when: L1 = 10 mH L2 = 15 mH L2= 7.5 mH Ans: 32.5 mH

Before starting this module, you should be able to:  

When you complete this module, you should be able to:

Describe how inductors oppose any change in current flowing through them. Cite the units of measure for inductance.

    

Cite the inverse formulas for total inductance of a parallel circuit. Explain how total inductance decreases with each inductor that is added in parallel. Calculate the total inductance of a parallel circuit, given the values of individual inductors. Describe how the voltage is the same across each inductor. Describe how the current in each branch reponds to changes in applied voltage.

Topic 3-3.1 Formula for Total Inductance in a Parallel Circuit A parallel inductor circuit consists of two or more inductors connected so that each one is in a separate branch. The total inductance of a parallel inductor circuit is less than the value of the smallest inductor. The total inductance of a parallel inductor circuit is found by applying one of the reciprocal

The total inductance of exactly two inductors connected in parallel can be determined by using the product-over-sum rule:

formulas: or

where: LT = the total inductance in henries (H) L1, L2, L3, Ln = the value of the individual inductances in henries (H)

Determine the total inductance of this circuit when L1 = 30 mH and L2 = 40 mH. Ans: 17.1 mH

Calculate the total inductance of this circuit when: L1 = 750 mH L2 = 1.5 mH Ans: 500 mH

Determine the total inductance of this circuit when: L1 =750 mH L2 = 1.0 mH L3 = 2.2 mH Ans: 359 mH

Unit 4 Direct-Current RL Circuits

Lesson 4-2 RL Time Constant

\Before starting this module, you should be able to:   

Explain self-inductance. Define inductance in terms of induced voltage. Cite the units of measure for inductance.

When you complete this module, you should be able to:   

Note: User your browser's BACK feature to return to this point.     

L/R Time Constant The time constant of a series RL circuit equal to the value of inductance divided by the resistance: T=L/R

where T = time constant in seconds L = inductance in henries R = resistance in ohms

Describe the equation for determining the time constant of a series RL circuit. Calculate the L/R time constant of a circuit. Explain the significance of the percentage value 63.2% while current is building through an RL circuit. Explain why the build-up current of an inductor reaches its steady state the end of 5 time constants. Calculate the build-up current through an inductor after a given number of time constants. Explain the significance of the percentage value 63.2% while current is decaying through an RL circuit. Explain why the decaying current of an inductor reaches a steady state at the end of 5 time constants. Calculate the amount of decay current through an inductor after a given number of time constants.

The time constant for an RL circuit is nothing more than the value of the inductor divided by the value of the resistor.. You will also find the Greek letter  (tau) used as the math symbol for time constant. For instance:  =L/R

What is the time constant of a series RL circuit where R = 1 k and L = 1 mH? Ans: 1 ms

Top RL Build-Up Curve

RL Current Build-Up Table Current through the inductor in an RL circuit does not increase at a steady rate. Rather, the rate of increase is rapid at first, but then slows as it reaches the maximum level.

During each time constant, the current builds 63.2% of the remaining distance to the maximum current level.

T

iL

0

0

1

0.632 x Vs/R

2

0.865 x Vs/R

3

0.950 x Vs/R

4

0.981 x Vs/R

5

0.992 x Vs/R

This table shows how to calculate the build-up current through an inductor at the end of each time constant. T = number of time constants that have passed in seconds Vs = voltage of the DC source in volts R = value of the series resistor in ohms

Inductor current build-up is considered complete at the end of 5 time constants.

The ratio Vs / R is actually an expression of Ohm's Law for maximum current through the circuit. You can always replace the ratio Vs / R with Imax. Vs / R = Imax Step 1

Vs = 6 V R = 100  L = 100 mH

T=L/R T = 100 mH / 100  T = 1 ms Step 2 At the end of two time constants:

Ans: T = 1 ms, iL = 51.9 mA

iL = 0.865 x Vs/R

iL = 51.9 mA Top RL Decay Curve

RL Current Decay Table Inductor current does not drop off at a steady rate. Rather, the rate of current decay is discharge is rapid at first, but slows considerably as the charge approaches zero.

During each time constant, the current decays 63.2% of the remaining distance to the minimum current level.

Inductor current decay is considered complete at the end of 5 time constants.

T

iL

0

Vs/R

1

0.368 x Vs/R

2

0.135 x Vs/R

3

0.05 x Vs/R

4

0.019 x Vs/R

5

0.008 x Vs/R

This table shows how to calculate the decay current through an inductor at the end of each time constant. T = number of time constants that have passed in seconds Vs = voltage of the DC source in volts R = value of the series resistor in ohms Step 1 T=L/R T = 1.2 s

The steady-state maximum current through a 1.2 H inductor is 12 A. When this inductor is switched from the power source to a 1 W resistor, what is the current at the end of 3 time constants.

Step 2 At the end of 3 time constants: Ans: iL = 600 mA iL = 0.05 x Vs / R, but remember that Vs / R = Imax: iL = 0.05 x 12 A iL = 600 mA

Lesson 5-2 Transformer Ratios

Before starting this module, you should be able to:

When you complete this module, you should be able to:  



    

Voltage-Turns Ratio

Cite the equation for the voltage-turns ratio of a transformer. Describe exactly how the turns ratio of a transformer is related to its voltage ratio. Explain the meaning of the terms step-up transformer and step-down transformer. Cite the equation for the current-turns ratio of a transformer. Describe exactly how the turns ratio of a transformer is rlated to its current ratio. Cite the equation for the current-voltage ratio of a transformer. Describe exactly how the current ratio of a transformer is related to its voltage ratio.

The voltage ratio of a transformer is directly related to the turns ratio.

A step-up transformer is one whose secondary voltage (VS) is greater than its primary voltage (VP). This kind of transformer "steps up" the voltage applied to it. where: VS = secondary voltage VP = primary voltage NS = number of turns in the secondary winding NP = number of turns in the primary winding

A step-down transformer is one whose secondary voltage (VS) is less than its primary voltage (VP). This kind of transformer "steps down" the voltage applied to it.

The RMS voltage on the primary of a transformer is 120 V, and the RMS voltage on the secondary is 12 V.

What is the turns ratio of this transformer? Is it a step-up or a step-down transformer?

Ans: The turns ratio is 10:1. It is a step-down transformer.

The situation portrayed in this example is a realistic one. It is usually very easy to use a voltmeter to determine the primary and secondary voltages of a transformer. Otherwise, you usually have to consult the manufacturer's data sheets to determine the turns ratio.

What is the secondary voltage of a 6.5:1 step-up transformer when 48 V is applied to the primary winding?

This type of question can be easily handled by this version of the voltage/turns-ratio equation:

Ans: 312 V

Top Voltage-Turns Ratio

The current ratio of a transformer is inversely related to the turns ratio.

where: IP = primary current IS = secondary current NS = number of turns in the secondary winding NP = number of turns in the primary winding

The RMS current on the primary of a transformer is 120 mA, and the RMS current on the secondary is 2.4 A. What is the turns ratio (NS/NP) of this transformer?

It is usually easier to determine the turns ratio of a transformer by measuring and comparing the primary and secondary voltages than by the currents.

Ans: 20:1

A certain transformer has a turns ratio (NS:NP) of 2.5:1. What is the primary current when the secondary current is 55 mA?

This type of question can be easily handled by this version of the voltage/turns-ratio equation:

Ans: 1.38 A Or if the situation calls for determining secondary current, given the turns ratio and primary current, use this form of the equation:

Top Voltage-Current Ratio

This equation clearly shows the inverse relationship between the voltages and currents of a transformer:  

where: VS = secondary voltage VP = primary voltage IP = primary current IS = secondary current

A transformer that steps up voltage will step down current. A transformer that steps down voltage will step up current.

The voltage-current transformer ratio is derived directly from the fact that the power dissipated in the primary of a transformer is (in theory) equal to the power dissipated in the secondary: PS = PP

A certain step-down transformer has a Vs / Vp ratio of 10. What is the amount of primary current when the secondary current is 100 mA? Ans: 1 A

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Unit 6 Inductive Reactance

Lesson 6-1 Introduction to Inductive Reactance

Before starting this module, you should be able to:  

When you complete this module, you should be able to:  

Define inductance and describe its schematic symbol. Cite the units of measure for inductance.

Define inductive reactance. Describe the effect that inductive reactance has upon the amount of current flowing in an AC circuit.

Note: Use the BACK function of your browser to return to this page. Inductive Reactance

The amount of inductive reactance in a circuit is proportional to:  

Inductive reactance is the opposition to AC current flow that is caused by the presence of an inductor in the circuit.

 

The symbol for inductive reactance is XL. The units of measure for inductive reactance is ohms, W.

Applied frequency, f Value of the inductor, L

Inductive reactance is an AC version of resistance. In fact, you can use Ohm's Law by substituding XL for R: VL = ILXL where: VL is the voltage across the inductor in volts IL is the current through the inductor in amperes XL is the amount of inductive reactance in ohms

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Lesson 6-2 The Inductive Reactance Formula

Before starting this module, you should be able to:  

When you complete this module, you should be able to: 

Define inductive reactance. Describe the effect that inductive reactance has upon the amount of current flowing in an AC circuit.



Cite the equation for determining the value of inductive reactance, given the values of applied frequency and inductance. Solve the equation, given two of the three variables.

Note: Use the BACK function of your browser to return to this page. The equation for calculating the amount of inductive reactance in an ac circuit is given by:

The equation, XL = 2fL , demonstrates the relationship between inductive reactance (XL), the frequency (f) of the waveform applied to the circuit, and the value of the inductance (L).

XL = 2fL where:

The amount of inductive reactance (XL) changes proportionally with the applied frequency (f):

XL = inductive reactance ohms (W) f = frequency in hertz (Hz) L = inductance in henries (H)

 

Increasing the value of f causes XL to increase. Decreasing the value of f causes XL to decrease.

The amount of inductive reactance (XL) changes proportionally with the value of inductance (L):

 

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Increasing the value of L causes XL to increase. Decreasing the value of L causes XL to decrease.

Use the basic equation: What is the value of inductive reactance for an 0.1 H coil that is operating at 1 kHz?

XL = 2fL

Ans: 628 

Use this form of the basic equation: What value inductor is required for producing an inductive reactance of 10  at 1.8 kHz?

L = XL / (2f )

Ans: 88.5 H

Use this form of the basic equation: At what frequency will a 150 mH inductor have an inductive reactance of 150? Ans: 159 Hz

f = XL / (2L)

Lesson 6-3 Series and Parallel XL Before starting this module, you should be able to:   

When you complete this module, you should be able to:

Define inductive reactance. Describe the effect that inductive reactance has upon the amount of current flowing in an AC circuit. Cite the equation for determining the value of inductive reactance, given the values of applied frequency and inductance.

   

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Explain the meaning of each term in the equation for total inductive reactance in a series circuit. Calculate the total inductive reactance of a series circuit, given the values of the individual reactanaces. Explain the meaning of each term in the equation for total inductive reactance in a parallel circuit. Calculate the total inductive reactance of a parallel circuit, given the values of the individual reactanaces.

Topic 6-3.1 Inductive Reactances in Series Series Inductive Reactances

The total inductive reactance of a series XL circuit is equal to the sum of the individual reactances.

XLT = XL1 + XL2 + XL3 + ... + XLn where: XLT = total inductive reactance XL1 , XL2 , XL3 , XLn = values of the individual reactances

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What is the total inductive reactance of this circuit when X L1 = 150  and XL2 = 75 ? Ans: 225  Top

Topic 6-3.2 Inductive Reactances in Parallel

Parallel Inductive Reactances

The procedure for finding the total inductive reactance of a parallel inductor circuit is identical to finding the total resistance of a parallel resistor circuit. The total reactace of two inductors in parallel can be found by applying the product-oversum formula:

Use one of these inverse equations to determine the total inductive reactance of a parallel inductor circuit:

or

Where: XLT = total inductive reactance XL1 , XL2 , XL3 , XLn = values of the individual reactances

What is the total inductive reactance of this circuit when X L1 = 150  and XL2 = 200 ? Ans: 85.7 

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Lesson 6-4 XL and Ohm’s Law

Before starting this module, you should be able to:  



When you complete this module, you should be able to:  

Describe Ohm's Law for resistance, and explain how the value of resistance affects the amount of current flowing through a DC circuit. Cite the equation for determining the value of inductive reactance, given the values of applied frequency and inductance.

 

Cite Ohm's Law for inductive reactance. Explain how Ohm's Law for XL is similar to Ohm's Law for R. Use Ohm's Law to solve for voltage, current, or inductive reactance for an inductor. Solve Ohm's Law for inductance, given values of f and L rather than XL.

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Topic 6-4.1 Basic Application of Ohm's Law for XL

Just as resistance opposes the flow of current through a resistor, inductive reactance opposes the flow of current through an inductor.

Ohm's Law for a resistor: V = IR Ohms' Law for an inductor: VL = ILXL

Ohm's Law applies directly to an inductor: VL = ILXL where: VL = voltage across the inductor IL = current through the inductor XL = inductive reactance

Solution: For a certain inductor, IL = 20 mA and XL = 420 . What is the voltage across this inductor?

This is a straightforward application of Ohm's Law for XL.

Ans: 8.4 V VL = ILXL

Solution: What is the current through an inductive reactance of 12 kW when the voltage across it is 12.6 V?

Given the values for XL and VL, use this form of Ohm's Law to solve for IL:

Ans: 1.05 mA IL= VL / XL

Solution: The current through an inductor is 250 mA when 16 V is dropped across it. What is the value of XL?

Given the values for IL and VL, use this form of Ohm's Law to solve for XL:

Ans: 64 W XL= VL / IL

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Topic 6-4.2 Ohm's Law when XL is Not Known R XL = Basic definition of E 2fL inductive reactance. A S VL = Ohm's Law for inductive reactance. O ILXL N I It follows that: N VL =2fL IL G

There aren't many electronics labs that are equipped to measure the X L of an inductor directly. Instead, XL is usually calculated from the value of the inductor and the frequency of the sine waveform applied to it. So when you want to use Ohm's Law to determine the current of voltage across an inductor, you must first calculate the value of X L, then solve for the current or voltage. Step 1: Calculate XL = 2fL Step 2: Substitute the calculated value of XL into Ohm's Law, VL = ILXL. Or you can combine the two equations to produce the single equation as shown on the left.

Lesson 6-5 Current and Voltage in XL Circuits

Before starting this module, you should be able to:   

When you complete this module, you should be able to:

Explain how Ohm's Law for XL is similar to Ohm's Law for R. Use Ohm's Law to solve for voltage, current, or inductive reactance for an inductor. Solve Ohm's Law for inductance, given values of f and L rather than XL.

 

Describe the fact that the current through an inductor always lags the voltage across the inductor by 90 degrees. Sketch a vector diagram showing how the current lags the voltage.

 Changing the amount of voltage applied to an inductor causes a R corresponding change in current through the inductor. E  Due to the property of self-inductance, however, changes in A inductor current always lag behind the changes in applied voltage. S  When the applied voltage is a sinusoidal waveform, the voltage is changing constantly and the current is constantly lagging behind. O N I Note: N Unless stated otherwise, discussions of inductive reactance in AC circuits G assume a sinusoidal voltage source.

The current through an inductor lags the voltage applied to the inductor by 90°

It is also correct to say that the voltage applied to an inductor leads the current through the inductor by 90° The 90° phase difference between current and voltage of an inductor applies only to ideal inductors—inductors that have no internal resistance. The internal resistance of real-world inductors causes the phase difference to be something less than 90°.

The phase difference between inductor current and voltage can also be shown with a vector diagram:

Lesson 6-6 Power in XL Circuits Before starting this module, you should be able to:   

When you complete this module, you should be able to: 

Sketch voltage, current, and power sine waveforms on the same axis. Explain why the power waveform is always positive as long as current and voltage are in phase. Describe the fact that the current through an inductor always lags the voltage across the inductor by 90 degrees.

 

Describe how AC power is absorbed by an inductor for one-quarter cycle, then returned to the circuit during the next quarter cycle. Explain the meaning of each term in the equation for instantaneous power in an AC inductor circuit. Describe the differences between power in a resistor circuit and apparent power in an inductor circuit.

Note: Use the BACK function of your browser to return to this page. In an AC circuit, an inductor alternately absorbs and returns power to the circuit. AC power that is used for building up the magnetic lines of force is subsequently returned to the circuit as the magnetic lines of force collapse.

The shaded green areas in this diagram show how power is absorbed and returned to the circuit. 

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The green areas above the baseline ( + levels) represent power that is absorbed by the inductor. The green areas below the baseline (- levels) representpower that is returned to the circuit.

Top INSTANTANEOUS POWER The equation for determining the power at any given instant is: p = -½Ppsin2 where p = instantaneous power Pp = peak power in watts  = angle in radians

In a purely inductive circuit, the amount of power absorbed by building up the inductor's magnet field is exactly equal to the amount of power returned to the circuit when the field collapses. The average power dissipation in a purely inductive circuit is zero. The inductor absorbs power for onequarter of the applied AC cycle and returns it to the circuit during the next quarter cycle.

Top APPARENT POWER Power in a purely resistive circuit is found by multiplying the RMS voltage times the RMS current: P = VI Apparent power in a purely inductive circuit is found the same way, but the name, symbol, and units of measure are slightly different. S = VI where S = apparent power in volt-ampere reactive, VAR V = voltage across the inductor I = current through the inductor

It is important to remain aware of the differences in terminology, symbols, and units for power in purely resistive and purely inductive circuits. Resistor Circuit Name: power Symbol: P Unit: watt, W Equation: P = VI

Inductor Circuit Name: apparent power Symbol: S Unit: volt-ampere reactive, VAR Equation: S = VI